Identify from the following the correct set of thermodynamic conditions for the reaction to be spontaneous at all temperatures.

$A$ spontaneous process is impossible if ........

The maximum work (in $kJ \, mol^{-1}$) that can be derived from the complete combustion of $1 \, mole$ of $CO$ at $298 \, K$ and $1 \, atm$ is:
[Standard enthalpy of combustion of $CO = -283.0 \, kJ \, mol^{-1}$; standard molar entropies: $S_{O_2} = 205.1 \, J \, K^{-1} \, mol^{-1}$,$S_{CO} = 197.7 \, J \, K^{-1} \, mol^{-1}$,$S_{CO_2} = 213.7 \, J \, K^{-1} \, mol^{-1}$]

At constant pressure and temperature,the direction of any spontaneous chemical reaction is towards the state where there is a decrease in

Calculate the amount of work done during isothermal expansion of a gas from a volume of $4 \ dm^{3}$ to $6 \ dm^{3}$ against a constant external pressure of $3 \ atm$ (in $J$)?

The standard state Gibbs free energies of formation of $C$ (graphite) and $C$ (diamond) at $T = 298 \ K$ are:
$\Delta_f G^0[C(\text{graphite})] = 0 \ kJ \ mol^{-1}$
$\Delta_f G^0[C(\text{diamond})] = 2.9 \ kJ \ mol^{-1}$
The standard state means that the pressure should be $1 \ bar$,and the substance should be pure at a given temperature. The conversion of graphite [$C$ (graphite)] to diamond [$C$ (diamond)] reduces its volume by $2 \times 10^{-6} \ m^3 \ mol^{-1}$. If $C$ (graphite) is converted to $C$ (diamond) isothermally at $T = 298 \ K$,the pressure at which $C$ (graphite) is in equilibrium with $C$ (diamond) is:
[Useful information: $1 \ J = 1 \ kg \ m^2 \ s^{-2} ; 1 \ Pa = 1 \ kg \ m^{-1} \ s^{-2} ; 1 \ bar = 10^5 \ Pa$ ] (in $bar$)